A cube has an edge of 2 feet. The edge is increasing at the rate of 5 feet per minute. Express the volume of the cube as a function of m, the number of minutes elapsed.

Answer:

The correct option is  A

Step-by-step explanation:

From the question we are told that

    The  population is  [tex]\mu = 6.6[/tex]

     The level of significance is [tex]\alpha = 5\% = 0.05[/tex]

      The sample data is  9.0, 7.3, 6.0, 8.8, 6.8, 8.4, and 6.6 pounds

The Null hypothesis is [tex]H_o : \mu = 6.6[/tex]

 The Alternative hypothesis is  [tex]H_a : \mu > 6.6[/tex]

The critical value of the level of significance obtained from the normal distribution table is

                       [tex]Z_{\alpha } = Z_{0.05 } = 1.645[/tex]

Generally the sample mean is mathematically evaluated as

      [tex]\=x = \frac{\sum x_i }{n}[/tex]

substituting values

      [tex]\=x = \frac{9.0 + 7.3 + 6.0+ 8.8+ 6.8+ 8.4+6.6 }{7}[/tex]

      [tex]\=x = 7.5571[/tex]

The standard deviation is mathematically evaluated as

           [tex]\sigma = \sqrt{\frac{\sum [ x - \= x ]}{n} }[/tex]

substituting values

          [tex]\sigma = \sqrt{\frac{ [ 9.0-7.5571]^2 + [7.3 -7.5571]^2 + [6.0-7.5571]^2 + [8.8- 7.5571]^2 + [6.8- 7.5571]^2 + [8.4 - 7.5571]^2+ [6.6- 7.5571]^2 }{7} }[/tex][tex]\sigma = 1.1774[/tex]

Generally the test statistic is mathematically evaluated as

            [tex]t = \frac{\= x - \mu } { \frac{\sigma }{\sqrt{n} } }[/tex]

substituting values

           [tex]t = \frac{7.5571 - 6.6 } { \frac{1.1774 }{\sqrt{7} } }[/tex]

            [tex]t = 1.4274[/tex]

Looking at the value of  t and  [tex]Z_{\alpha }[/tex]   we see that [tex]t < Z_{\alpha }[/tex] hence we fail to reject the null hypothesis

  What this implies is that there is no sufficient evidence to state that the sample data show as significant increase in the average birth rate

The conclusion is that the mean is  [tex]\mu = 6.6 \ lb[/tex]

Answer:

There is not enough information to the problem.

The things we know are:

Zhi wants to draw a king of clubs or an ace of clubs.

There is a 52 card deck, but we do not know which cards are in play.

Now, the probability of drawing two specific cards out of a deck of 52 cards is equal to:

P = 2/52 = 0.038.

Now, suppose that there are X cards in game, and Zhi knows the cards (and he can see that the king of clubs and the ace of clubs are not in the table, so the must be on the card pile), now the probability is bigger, because now we want to draw two specific cards out of 52 - X cards, so the probability now is:

p = 2/(52 - X)

and 52 - X is smaller than 52, so the denominator is smaller, which means that the probability is actually larger.

Answer:

31%

Step-by-step explanation:

Since the two events, drawing a face card and drawing an ace card, are non-overlapping, we can use the following formula:

P left parenthesis A c e space o r space F a c e right parenthesis equals P left parenthesis A c e right parenthesis plus P left parenthesis F a c e right parenthesis equals 4 over 52 plus 12 over 52 equals 16 over 52 equals 4 over 13 equals space 0.308 space o r space 31 percent sign

Answer:

you could stand at 5.0 ft and still be completely in the shadow of the tree

Step-by-step explanation:

From the diagram attached below;

We consider;

[tex]\overline {BC}[/tex] to be the height of the tree and [tex]\overline {DE}[/tex] to be the height of how tall you could be and still be completely in the shadow of the tree.

∠D = ∠B = 90°

Also;

ΔEAD = ΔBAC   (similar triangles)

Therefore, their sides will also be proportional

i.e

[tex]\dfrac{\overline {DE}}{ \overline {BC}}= \dfrac{\overline{AD}}{ \overline{AC}}[/tex]

[tex]\dfrac{x}{ 45}= \dfrac{225-220}{225}[/tex]

[tex]\dfrac{x}{ 45}= \dfrac{25}{225}[/tex]

By cross multiply

225x = 45 × 25

[tex]x = \dfrac{45 \times 25}{225}[/tex]

[tex]x = \dfrac{1125}{225}[/tex]

x = 5.0 ft

Therefore, you could stand at 5.0 ft and still be completely in the shadow of the tree

Answer:

√5

Step-by-step explanation:

[tex](2 ,-5) = (x_1,y_1)\\(3,-7)=(x_2,y_2)\\\\d = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\ \\d = \sqrt{(3-2)^2 +(-7-(-5))^2}\\ \\d = \sqrt{(1)^2+(-7+5)^2}\\ \\d = \sqrt{(1)^2 + (-2)^2}\\ \\d = \sqrt{1 +4}\\ \\d = \sqrt{5}[/tex]

Answer:

angle B is 62 Degress angle A is 87 degress D is 87 degress C is 28 degress.

Step-by-step explanation:

I am in geometry btw so i know this stuff and 65 plus 28 is 93 and 180 -93 is 87 so a is 87 and d is 87 too becuase of vertical angles and b is 62 becuase 90 -28 is 62 and c is 28 becuase of vertical angles your wellcome kid good luck!!!!

Answer:

1/9

Step-by-step explanation:

easy 2/3 is equivalent to 6/9. So there is 1/9 of a pint left

Answer: When the addition of two or more equations leads to the elimination of one of the variables.

Step-by-step explanation:

When we have a system of equations, the addition method seems to be useful only when adding the equations will lead to the elimination of one of the variables:

An example of this can be, for the variables x and y:

3*x + x*y - 2*y = 3

x^2 + x*y - 2y = 42

now we can "add" (actually subtract) the equations and get (eq2 minus eq1)

(x^2 + x*y - 2y) - (3*x + x*y - 2*y ) = 42 - 3

x^2 - 3*x = 39

x^2 - 3*x - 39 = 0

And now we can solve it for x, and then find the value of y.

Answer:

nd is c

Step-by-step explanation:

Answer:  sqrt(27.2) =approx 5.215

Step-by-step explanation:

The distance between 2 points can be calculated using Phitagor theorem

L= sqrt( (x1-x2)²+(y1-y2)²)

Where x1, y1 are the coordinates of the first point and  x2, y2 are the coordinates of the 2-nd point.

L=sqrt((3.1-2.7)²+(0.3-(-4.9))²)= sqrt(0.4²+5.2²)=sqrt(27.2) - this is exact answer.

sqrt(27.2)=5.21536...=approx 5.215

julissa gave equal oranges in 6 apartments

she gave each apartment 5 oranges

so total no. of oranges are = 6×5 = 30

Answer:

D. 30

Step-by-step explanation:

Answer:

(A) The shape of the distribution of the sample mean is bell-shaped.

(B) The standard error of the distribution of the sample mean is 1.1.

(C) The proportion of the samples that have a mean useful life of more than 36 hours is 0.1814.

(D) The proportion of the sample that has a mean useful life greater than 34.5 hours is 0.6736.

(E) The proportion of the sample that has a mean useful life between 34.5 and 36.0 hours is 0.4922.

Step-by-step explanation:

We are given that Power +, Inc. produces AA batteries used in remote-controlled toy cars. The mean life of these batteries follows the normal probability distribution with a mean of 35.0 hours and a standard deviation of 5.5 hours.

As a part of its quality assurance program, Power +, Inc. tests samples of 25 batteries.

Let [tex]\bar X[/tex] = sample mean life of these batteries

(A) The shape of the distribution of the sample mean will be bell-shaped because the sample mean also follows the normal distribution as it is taken from the population data only.

(B) The standard error of the distribution of the sample mean is given by;

            Standard error =  [tex]\frac{\sigma}{\sqrt{n} }[/tex]

Here, [tex]\sigma[/tex] = standard deviation = 5.5 hours

         n = sample of batteries = 25

So, the standard error =  [tex]\frac{5.5}{\sqrt{25} }[/tex]  = 1.1.

(C) The z-score probability distribution for the sample mean is given by;

                               Z  =  [tex]\frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } }[/tex]  ~ N(0,1)

where, [tex]\mu[/tex] = population mean life of battery = 35.0 hours

            [tex]\sigma[/tex] = standard deviation = 5.5 hours

            n = sample of batteries = 25

Now, the proportion of the samples that will have a mean useful life of more than 36 hours is given by = P([tex]\bar X[/tex] > 36 hours)

       P([tex]\bar X[/tex] > 36 hours) = P( [tex]\frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } }[/tex] > [tex]\frac{36-35}{\frac{5.5}{\sqrt{25} } }[/tex] ) = P(Z > 0.91) = 1 - P(Z [tex]\leq[/tex] 0.91)

                                                               = 1 - 0.8186 = 0.1814

(D) The proportion of the samples that will have a mean useful life of more than 34.5 hours is given by = P([tex]\bar X[/tex] > 34.5 hours)

       P([tex]\bar X[/tex] > 34.5 hours) = P( [tex]\frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } }[/tex] > [tex]\frac{34.5-35}{\frac{5.5}{\sqrt{25} } }[/tex] ) = P(Z > -0.45) = P(Z [tex]\leq[/tex] 0.45)

                                                                    = 0.6736

(E) The proportion of the samples that will have a mean useful life between 34.5 and 36.0 hours is given by = P(34.5 hrs < [tex]\bar X[/tex] > 36 hrs)

     P(34.5 hrs < [tex]\bar X[/tex] < 36 hrs) = P([tex]\bar X[/tex] < 36 hrs) - P([tex]\bar X[/tex] [tex]\leq[/tex] 34.5 hrs)

     P([tex]\bar X[/tex] < 36 hours) = P( [tex]\frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } }[/tex] < [tex]\frac{36-35}{\frac{5.5}{\sqrt{25} } }[/tex] ) = P(Z < 0.91) = 0.8186

     P([tex]\bar X[/tex] [tex]\leq[/tex] 34.5 hours) = P( [tex]\frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } }[/tex] [tex]\leq[/tex] [tex]\frac{34.5-35}{\frac{5.5}{\sqrt{25} } }[/tex] ) = P(Z [tex]\leq[/tex] -0.45) = 1 - P(Z [tex]\leq[/tex] 0.45)

                                                                    = 1 - 0.6736 = 0.3264                              

Therefore, P(34.5 hrs < [tex]\bar X[/tex] < 36 hrs) = 0.8186 - 0.3264 = 0.4922.

Answer:

hello your question is incomplete below is the complete question

Find the minimum sample size n needed to estimate μ For the given values of​ c, σ​, and E. c=0.98​, σ=6.5​, and E=22 Assume that a preliminary sample has at least 30 members.

Answer : 48

Step-by-step explanation:

Given data:

E = 2.2,

std ( σ ) = 6.5

c ( level of confidence ) = 0.98

To find the minimum sample size

we have to first obtain the value of  [tex]Z_{a/2}[/tex]  

note : a can be found using this relation :

( 1 - a ) = 0.98 ----- equation 1

a = 1 - 0.98 = 0.02

hence:  a/2 = 0.01

This means that P( Z ≤ z ) = 0.99  the value of z can be found using the table of standard normal distribution. from the table the value of z = 2.33

P( Z ≤ 2.33 ) = 0.99

To obtain the sample size n

[tex]n = (\frac{std*z}{E} )^{2}[/tex]

n = [tex](\frac{6.5*2.33}{2.2} )^2[/tex] =  (6.88409)^2

Therefore n ≈ 48

Answer:

[tex]\huge\boxed{4 ( 20 + 21)}[/tex]

Step-by-step explanation:

4(20) + 84

Resolve Parenthesis

80 + 84

Taking 4 common as both are the multiples of 4

4 ( 20 + 21)

Answer:

The  98% confidence interval

                         [tex]0.1884 < p < 0.2876[/tex]

The confidence interval contains  20.4%

Step-by-step explanation:

From the question we are told that

    The sample size is  n =  400

The number that replied that they were considering business as a major [tex]x = 95[/tex]

  The  sample proportion is mathematically  evaluated as

          [tex]\r p = \frac{95}{400}[/tex]

         [tex]\r p = 0.238[/tex]

Given that the confidence level 98% then the level of significance is evaluated as

      [tex]\alpha = 100 - 98[/tex]

     [tex]\alpha = 2 \%[/tex]

     [tex]\alpha = 0.02[/tex]

Next we obtain the critical value of [tex]\frac{ \alpha }{2}[/tex]  from the normal distribution table is  

       [tex]Z_{\frac{ \alpha }{2} } = 2.33[/tex]

  Generally the margin of error is mathematically represented as

       [tex]E = Z_{\frac{ \alpha }{2} } * \sqrt{ \frac{ p (1 - p )}{n} }[/tex]  

       [tex]E = 2.33 * \sqrt{ \frac{ 0.238 (1 - 0.238 )}{400} }[/tex]

        [tex]E = 0.0496[/tex]

The  98%  confidence interval is mathematically represented

   [tex]\r p - E < p < \r p + E[/tex]

 =>    [tex]0.238 - 0.0496 < p <0.238 + 0.0496[/tex]

=>      [tex]0.1884 < p < 0.2876[/tex]

Answer:

Step-by-step explanation:

[tex]27^{\frac{2}{3}}\\\mathrm{Factor\:the\:number:\:}\:27=3^3\\=\left(3^3\right)^{\frac{2}{3}}\\\mathrm{Apply\:exponent\:rule}:\\\\\quad \left(a^b\right)^c=a^{bc},\:\quad \:a\ge 0\\\\\left(3^3\right)^{\frac{2}{3}}=3^{3}\times \frac{2}{3}}\\\\3\=times \frac{2}{3}=2\\\\=3^2 \\\\=9[/tex]

[tex]27^{2/3}=(3^3)^{2/3}=3^2=9[/tex]

You said    - 1/3 - 3/5 x  =  1/2

Multiply each side by 3 :

- 1 - 9/5 x  =  3/2

Multiply each side by 5 :

- 5 - 9x  =  15/2

Multiply each side by 2 :

- 10 - 18x = 15

Add 10 to each side :

- 18x  =  25

Divide each side by -18 :

x = - 25/18

or  x = - 1 and 7/18 (same thing)

Answer:

Below

Step-by-step explanation:

The two given expressions are:

● √(2p-7) = 3

● 7√(3q-1) = 2

We are told to evaluate p+q^2

To do that let's find the values of p and q^2

■■■■■■■■■■■■■■■■■■■■■■■■■■

Let's start with p.

● √(2p-7) = 3

Square both sides

● (2p-7) = 3^2

● 2p-7 = 9

Add 7 to both sides

● 2p-7+7 = 9+7

● 2p = 16

Divide both sides by 2

● 2p/2 = 16/2

● p = 8

So the value of p is 8

■■■■■■■■■■■■■■■■■■■■■■■■■■

Let's find the value of q^2

● 7√(3q-1) = 2

Square both sides

● 7^2 × (3q-1) = 2^2

● 49 × (3q-1) = 4

● 49 × 3q - 49 × 1 = 4

● 147q - 49 = 4

Add 49 to both sides

● 147q -49 +49 = 4+49

● 147q = 53

Divide both sides by 147

● 147q/147 = 53/147

● q = 53/ 147

Square both sides

● q^2 = 53^2 / 147^2

● q^2 = 2809/21609

■■■■■■■■■■■■■■■■■■■■■■■■■

● p+q^2 = 8 +(2809/21609)

● p+q^2 = (2809 + 8×21609)/21609

● p+q^2 = 175681 / 21609

● p + q^2 = 8.129

Round it to the nearest unit

● p+ q^2 = 8

Answer:

Option 1: The terms in Pattern B is 4 times the corresponding terms of Pattern A

Step-by-step explanation:

Answer:

Pattern B has more then pattern A so option 2

Step-by-step explanation:

Answer:

24

Step-by-step explanation:

1+3=4

4 divided by 32 is 8

8 white ones

then 8-32 is 24

24 black ones

Answer:

a) No sufficient evidence to support the claim that the two machines produce rods with different mean diameters.

b) P-value is 0.80

c)  −0.3939 <μ< 0.4939

Step-by-step explanation:

Given Data:

sample sizes

n1 = 15

n2 = 17

sample means:

x1 = 8.73

x2 = 8.68

sample variances:

s1² = 0.35

s2² = 0.40

Hypothesis:

H₀ : μ₁ = μ₂

H₁ :  μ₁ ≠ μ₂

Compute the pooled standard deviation:

[tex]s_{p} = \sqrt{\frac{(n_{1} - 1)s_{1}^{2} + (n_{2} - 1)s_{2}^{2}}{n_{1} +n_{2} -2} }[/tex]

    [tex]= \sqrt{\frac{(15-1)0.35+(17-1)0.40}{15+7-2}}[/tex]

    [tex]= \sqrt{\frac{(14)0.35+(16)0.40}{30}}[/tex]

 [tex]= \sqrt{\frac{4.9+6.4}{30}}[/tex]

 [tex]= \sqrt{\frac{11.3}{30}}[/tex]

[tex]= \sqrt{0.376667}[/tex]

= 0.613732

= 0.6137

Compute the test statistic:

[tex]t = \frac{x_{1} -x_{2} }{s_{p} \sqrt{\frac{1}{n_{1} }+\frac{1}{n_{2} } } }[/tex]

[tex]= \frac{8.73-8.68}{0.6137\sqrt{\frac{1}{15}+\frac{1}{17} } }[/tex]

[tex]= \frac{0.05}{0.6137\sqrt{0.06667+0.05882} } }[/tex]

[tex]= \frac{0.05}{0.6137\sqrt{0.12549} } }[/tex]

[tex]= \frac{0.05}{0.6137(0.354246)} } }[/tex]

[tex]= \frac{0.05}{0.6137(0.354246)} } }[/tex]

= 0.05 / 0.217401

= 0.22999

t = 0.230

Compute degree of freedom:

df = n1 + n2 -2 = 15 + 17 - 2 = 30

Compute the P-value from table using df = 30

P > 2 * 0.40 = 0.80

P > 0.05 ⇒ Fail to reject H₀

Null hypothesis is rejected when P-value is less than or equals to level of significance. But here the P-value = 0.80 and level of significance = 0.05. So P-value is greater than significance level. Hence there is not sufficient evidence to support the claim that population means are different.

Construct a 95% confidence interval for the difference in mean rod diameter:

confidence = c = 95% = 0.95

α = 1 - c

  = 1 - 0.95

α = 0.05

Compute degree of freedom:

df = n1 + n2 -2 = 15 + 17 - 2 = 30

Compute [tex]t_{\alpha /2}[/tex] with df = 30 using table:

t₀.₀₂₅ = 2.042

Compute confidence interval:

= [tex](x_{1}-x_{2})-t_{\alpha/2} ( s_{p} )\sqrt{\frac{1}{n_{1} }+\frac{1}{n_{2} } }[/tex]

= (8.73 - 8.68) -  2.042 ( 0.6137 ) [tex]\sqrt{\frac{1}{15} +\frac{1}{17} }[/tex]

= 0.05 - 2.042 ( 0.6137 ) [tex]\sqrt{\frac{1}{15} +\frac{1}{17} }[/tex]

= 0.05 - 1.253175 [tex]\sqrt{0.06667+0.05882} } }[/tex]

= 0.05 - 1.253175 [tex]\sqrt{0.12549} } }[/tex]

= 0.05 - 1.253175 (0.35424))

= 0.05 - 0.443925

= −0.393925

= −0.3939

[tex](x_{1}-x_{2})+t_{\alpha/2} ( s_{p} )\sqrt{\frac{1}{n_{1} }+\frac{1}{n_{2} } }[/tex]

= (8.73 - 8.68) +  2.042 ( 0.6137 ) [tex]\sqrt{\frac{1}{15} +\frac{1}{17} }[/tex]

= 0.05 + 2.042 ( 0.6137 ) [tex]\sqrt{\frac{1}{15} +\frac{1}{17} }[/tex]

= 0.05 + 1.253175 [tex]\sqrt{0.06667+0.05882} } }[/tex]

= 0.05 + 1.253175 [tex]\sqrt{0.12549} } }[/tex]

= 0.05 + 1.253175 (0.35424))

= 0.05 + 0.443925

= 0.493925

= 0.4939

−0.3939 <μ₁ - μ₂< 0.4939

Answer:

It sold 14 cans boxes of food and 12 cans of food.

Step-by-step explanation:

The factor for the food cans depend upon every seven food boxes .So, the same no. of sets of food cans will be sold.

Let the no. of sets of food boxes be x.

According to the question,

6x+7x=26

13x=26

x=26/13

x=2

No. of food cans =6x=6×2=12 cans

No. of food boxes=7x=7×2=14 boxes

Please mark brainliest ,if it is truly the best ! Thank you!

Answer:

ax+182

Step-by-step explanation:

a*x+14*13

ax+182

Answer:

mean=87

median=87

Step-by-step explanation:

mean=sum of test score/number of subject

mean=79+91+93+85+86+88/6

mean=522/6

mean=87

Literal meaning of median is medium.

To find the number which lies in the medium, we must rearrange the number in ascending.

79, 91, 93, 85, 86, 88

79, 85, 86, 88, 91, 93

86+88/2=87

Hope this helps ;) ❤❤❤

Let me know if there is an error in my answer.

Answer:

[tex]f(-2)=33\\f(-1)=12\\f(0)=1\\f(1)=0\\f(2)=9[/tex]

Step-by-step explanation:

[tex]f(x)=5x^{2} -6x+1\\f(-2)=5(-2)^{2} -6(-2)+1\\f(-2)=5(4)+12+1\\f(-2)=20+13\\f(-2)=33[/tex]

[tex]f(x)=5x^{2}-6x+1\\f(-1)=5(-1)^{2} -6(-1)+1\\f(-1)=5(1)+6+1\\f(-1)=5+7\\f(-1)=12[/tex]

[tex]f(x)=5x^{2}-6x+1\\f(0)=5(0)^{2}-6(0)+1\\f(0)=5(0)-0+1\\f(0)=0+1\\f(0)=1[/tex]

[tex]f(x)=5x^{2}-6x+1\\f(1)=5(1)^{2}-6(1)+1\\f(1)=5(1)-6+1\\f(1)=5-5\\f(1)=0[/tex]

[tex]f(x)=5x^{2}-6x+1\\f(2)=5(2)^{2}-6(2)+1\\f(2)=5(4)-12+1\\f(2)=20-11\\f(2)=9[/tex]

Answer:

Daniel sold 54 and Jack sold 42

Step-by-step explanation:

D = number of tickets that Daniel sold

J = number of tickets that Jack sold

D + J = 96

D = 12+ J

Substitute the second equation into the first equation

12 + J + J = 96

Combine like terms

12 + 2J = 96

Subtract 12 from each side

2J = 84

Divide by 2

J = 42

D = J+12

D = 54

Daniel sold 54 and Jack sold 42

Answer:

Jack sold 42 & Daniel sold 54.

Step-by-step explanation:

96 - 12 = 84

84 / 2 = 42

Jack sold 42.

42 + 12 = 54

Daniel sold 54.

42 + 54 = 96

Answer:

maximum --> 62

median --> 46.5

upper quartile --> 60

lower quartile --> 37

minimum --> 32

Step-by-step explanation:

Forgive me on the explanation as I'm a bit rusty on these types of problems.

First, we need to put the set of numbers in order -->

from: 34, 37, 39, 32, 48, 45, 53, 62, 58, 61, 60, 41 -->

to: 32, 34, 37, 39, 41, 45, 48, 53, 58, 60, 61, 62

maximum = biggest number => thus, 62

median = middle number in a sense => (45+48)/2 => thus, 46.5

upper quartile = median over the median => thus, 60

lower quartile = median under the median => thus, 37

minimum = lowest number => thus, 32

And there we have our 5 answers.

Hope this helps!

Answer:

a) 40 dollars

b) 480 dollars

Step-by-step explanation:

Given the average property tax on a three bedroom home in a certain community modelled by the equation T(x) =20x²+40x+600, the rate at which the property tax is increasing with respect to time in 2008 can be derived by solving for the function T'(x) at x=0

T'(x) = 2(20)x¹ + 40x° + 0

T'(x) = 40x+40

At x = 0,

T'(0) = 40(0)+40

T'(0) = 40

Hence the property tax was increasing at a rate of 40dollars with respect to the initial year (2008).

b) There are 4 years between 2008 and 2012. To know how much that the tax change between the years 2008 and 2012, we will find T(4) - T(0)

Given T(x) =20x²+40x+600

T(4) =20(4)²+40(4)+600

T(4) = 320+160+600

T(4) = 1080 dollars

Also T(0) =20(0)²+40(0)+600

T(0) = 0+0+600

T(0)= 600 dollars

T(4) - T(0) = 1080 - 600

T(4) - T(0) = 480 dollars

Hence, the tax has changed by $480 between 2008 and 2012

Answer:

[tex]\boxed{\sf C. \ 10}[/tex]

Step-by-step explanation:

[tex]\sf The \ intersecting \ chord \ theorem \ states \ that \ the \ products[/tex]

[tex]\sf of \ the \ lengths \ of \ the \ line \ segments \ on \ each \ chord \ are \ equal.[/tex]

[tex]NH \times HT = MH \times HY[/tex]

[tex](x+20) \times 8=12 \times 20[/tex]

[tex]\sf Expand \ brackets \ and \ multiply.[/tex]

[tex]8x+160=240[/tex]

[tex]\sf Subtract \ 160 \ from \ both \ sides.[/tex]

[tex]8x+160-160=240-160[/tex]

[tex]8x=80[/tex]

[tex]\sf Divide \ both \ sides \ by \ 8.[/tex]

[tex]\displaystyle \frac{8x}{8} =\frac{80}{8}[/tex]

[tex]x=10[/tex]

The value of x is 10.

We have a circle and inside it two chords MY and NT intersect at point H.

We have to find the value of x in the figure.

According to the intersecting chord theorem, when two chords say AB and CD intersect at point O, then

AO x OB = CO x OD

Applying the chord intersecting theorem to the figure in the question, we get -

MH x HY = NH x HT

12 x 20 = (x+20) x 8

240 = 8x + 160

8x = 80

x = 10

Hence the value of x is 10.

To solve more questions on Circles and chords, visit the link below -

https://brainly.com/question/15568573

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Answer:

11/(x + 1) thus d: is the answer

Step-by-step explanation:

Simplify the following:

(11 (x + 8))/((x + 1) (x + 8))

(11 (x + 8))/((x + 1) (x + 8)) = (x + 8)/(x + 8)×11/(x + 1) = 11/(x + 1):

Answer: 11/(x + 1)

Complete Question

You look over the songs in a jukebox and determine that you like 18 of 59 songs.

(a) What is the probability that you like the next four songs that are played? (Assume a song cannot be repeated) Round to three decimal places as needed)

(b) What is the probability that you do not like the next four songs that are played? (Assume a song cannot be repeated.) Round to three decimal places as needed

Answer:

a

 [tex]P = 0.0067[/tex]

b

  [tex]Q = 0.222[/tex]

Step-by-step explanation:

From the question we are told that

    The  total number of songs is  [tex]n = 59[/tex]

    The  number of songs you liked is [tex]k = 18[/tex]

The probability that you like the next four songs that are played? (Assume a song cannot be repeated) is mathematically represented as

        [tex]P = \frac{ ^{k} C _4 }{ ^{n} C _4}[/tex]

=>     [tex]P = \frac{ ^{18} C _4 }{ ^{59} C _4}[/tex]

Now using a combination calculator

       [tex]P= \frac{ 3060}{ 455126}[/tex]

       [tex]P = 0.0067[/tex]

The probability that you do not like the next four songs that are played? (Assume a song cannot be repeated.) is mathematically evaluated as

     [tex]Q = \frac{ ^{n- k} C _4 }{ ^{n} C _4}[/tex]

=>  [tex]Q = \frac{ ^{59- 18} C _4 }{ ^{n} C _4}[/tex]

=>  [tex]Q = \frac{ ^{41} C _4 }{ ^{59} C _4}[/tex]

Now using a combination calculator

      [tex]Q = \frac{ 101270}{ 455126}[/tex]

      [tex]Q = 0.222[/tex]